I have this problem. Find the region of convergence of the following series of complex functions

$$ \sum_{n=1}^\infty \frac{2^n}{z^{2n}+1} $$

The progress I have made so far is that when n goes to infinity $f_n$ has a singularity everywhere in the border of the unit disc. So my intuition tells me that the series converges in the unit disc, although I don't know how to prove this.

  • $\begingroup$ maybe outside the unit disc? $\endgroup$ – user66081 Jan 29 '15 at 21:12
  • $\begingroup$ ((Accepted answer now (silently) changed. The current accepted post is mathematically sound but it does not answer the question, only half of it.)) $\endgroup$ – Did Feb 4 '15 at 7:00
  • $\begingroup$ I am sorry, Did. It's true it is not fully answered. I have unaccepted the answer now. As for the problem, I have not been able to do further progress myself. $\endgroup$ – Angel Alonso Feb 5 '15 at 19:11

Note that, for $|z|> \sqrt{2}$, you have $|z^{2n}+1|> |z|^{2n}-1>2^n-1$. Moreover (suppose $|z|> \sqrt{2} + \varepsilon$) you can show that $$\left| \frac{2^n}{z^{2n}+1} \right| < \rho^n $$ with suitable $\rho <1$ depending on $\varepsilon$, so the series converges (at least) in the unlimited domain $\{ |z| > \sqrt{2} \}$.

  • 1
    $\begingroup$ yes, here the bound with the absolute value is done correctly $\endgroup$ – Asier Calbet Jan 30 '15 at 16:44
  • $\begingroup$ This is mathematically sound but does not answer the question, only half of it. $\endgroup$ – Did Feb 4 '15 at 7:01

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